6
\$\begingroup\$

In my Java program, I need to operate on the sublists a of ArrayList or LinkedList often, like removing a sublist, comparing two sublists.

For ArrayList or LinkedList, I didn't find any good APIs to do these. My current implementation is as below. Basically, it takes an input like:

u1234 u1236 u1236 u2def u1236 u1236 u2def

and outputs the following:

u1234 ( ( u1236 ) * u2def ) *

Core function:

ArrayList<String> toRegex(ArrayList<String> tokenArray)
{
    /* check different length of continuous duplication */
    for(int len=1; len<=tokenArray.size()/2; len++) 
    {
        boolean match = false;
        /* given a length, scan for duplication */
        for(int i=0; i<tokenArray.size()-(2*len-1); i++) 
        {                
            ArrayList<String> first = new ArrayList<String>();
            ArrayList<String> second = new ArrayList<String>();
            for(int j=i; j<i+len; j++)
                first.add(tokenArray.get(j));
            for(int j=i+len; j<i+2*len; j++)
                second.add(tokenArray.get(j));                               

            if(isSpecial(first)) {       /* skip the cases */
                continue;
            }

            while(isIdentical(first, second))
            {
                match = true;
                for(int j=i+len; j<i+2*len; j++)
                    tokenArray.remove(i+len);
                if(i+2*len > tokenArray.size())
                    break;
                second.clear();
                for(int j=i+len; j<i+2*len; j++)
                    second.add(tokenArray.get(j));
            }

            if(match == true) {
                tokenArray.add(i, "(");
                tokenArray.add(i+1+len, ")");
                tokenArray.add(i+2+len, "*");
                i = i+3+len;
                match = false;
            }  
       }
    }

    return tokenArray;
}

Currently, the performance is not good. Could you find the bad design or inappropriate usage of data structures/APIs? What are the good common ways of operating sublists?

\$\endgroup\$
0

2 Answers 2

2
\$\begingroup\$

Yes, this doesn't look pretty. Consider converting the list to sets or using RegEx.

ArrayList<String> A = new ArrayList<String>();
A.add("Z");
A.add("Z");
A.add("C");
A.add("X");
A.add("Z");

Set<String> set = new HashSet<String>(A);

for(String temp:set)
    System.out.println(temp);           

Sublists:

List<String> subList = alist.subList(2, 4);

Use retainAll to get intersections:

listOne.retainAll(listTwo) ;  
boolean areEqualNotSamePosition = listeOne.size();
\$\endgroup\$
0
\$\begingroup\$
public static void main(final String[] args) {
        final String[] tabS = "u1234 u1236 u1236 u2def u1236 u1236 u2def "
            .split(" ");
            // The same thing can be done/adapted with a String[]
        final ArrayList<String> al = new ArrayList<>();
        final ArrayList<String> alTmp = new ArrayList<>();
        for (final String s : tabS) {
            al.add(s);
        }
        // removing need to begin by end
        for (int i = al.size() - 1; i >= 0; i--) {
            final String string = al.get(i);
            if (!alTmp.contains(string)) {
                alTmp.add(string);
            } else {
                al.remove(i);
            }
        }
        // inserting decoration if result have 2 or more object
        final int j = al.size();
        final String ending = ")*";
        if (j > 1) {
            for (int i = 1; i < j; i++) {
                al.add(1, "(");
            }
            for (int i = al.size() - 1; i > j; i--) {
                al.add(i, ending);
            }
            al.add(ending);
        }
        System.out.println(al.toString());
    }


Output : [u1234, (, (, u1236, )*, u2def, )*]
You can use a StringBuilder to remove ''' and make the output as you like.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.